Why is eπ√163e^{\pi \sqrt{163}} almost an integer?

The fact that Ramanujan’s Constant eπ163 is almost an integer (262537412640768743.99999999999925…) doesn’t seem to be a coincidence, but has to do with the 163 appearing in it. Can you explain why it’s almost-but-not-quite an integer in layman’s terms (I’m not a mathematician)?


This is quite a challenge to express in “layman’s terms”, but the
reason is that
is an integer where j is the j-function. When you substitute
(1+163)/2 into the q-expansion (see the wikipedia page)
of j, all terms save the first two are small, and the first two equal

The reason that this j-value is an integer is due to the quadratic
field Q(163) having
class number one, or equivalently that all positive-definite
integer binary quadratic forms of discriminant 163 are equivalent.

I’ll try to explain the connection with binary quadratic forms. Consider
a quadratic form
with a, b and c integers. I’ll only consider forms Q which are
primitive, so that a, b and c have no common factor >1,
and positive-definite, that is a>0 and the discriminant
D=b24ac<0. There is a notion of equivalence of quadratic forms,
and two primitive positive-definite forms Q and Q(x,y)=ax2+bxy+cy2
(necessarily also of discriminant D) are equivalent if and only if
For each possible discriminant there are only finitely many equivalence
classes. Thus we get a finite set of j-values for each discriminant, and
the big theorem is that they are the solutions of a monic algebraic equation
with integer coefficients. When there is only one class the equation has the
form xk=0 where x is an integer, and the j-value must be an integer.

My recommended reference for this is David Cox's book
Primes of the form x2+ny2. But these results appear towards
the end of this 350-page book.

Source : Link , Question Author : stevenvh , Answer Author : Robin Chapman

Leave a Comment